I'm curious about what happens if an explosive substance detonates in space. On Earth, I guess a good chunk of the energy released is carried away by shock waves in the atmosphere. But in space, the medium that supported the propagation of the shock wave on Earth is much more rarefied, so how does that work?

my instinct would be to check with experiment (which would be a very special kind of cool) but I doubt it's been done in any sort of large scale.
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Emilio PisantyAug 17 '12 at 2:03

I'm personally going to have a hard time setting up that experiment :-) - I was also thinking about supernovae "explosions" in astrophysics. Are those just shock waves that abruptly modify the properties of space like on Earth? Space still has a few atoms per square meter/kilometer of matter in it, right?
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FrankAug 17 '12 at 2:16

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@Frank: you actually do have shockwaves in the intergalactic medium due to supernovae, and the density peaks caused by this can cause new stars to form in the regions around the compressed region of the shockwaves, or it can cause stable stars to go supernova.
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Jerry SchirmerAug 17 '12 at 3:10

OK, I have no problem with that. But I'm wondering how it actually works. On Earth, in the atmosphere, I can see how the gas molecules push each other, propagating the shock wave (is that right?) - but if the intergalactic medium has 1 molecule per cubic kilometer, what are the odds of a molecule colliding with a neighbor to propagate the shock wave? Is that intergalactic shock wave then mostly electromagnetic energy?
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FrankAug 17 '12 at 3:25

@Frank: I suspect Jerry meant "interstellar" rather than "intergalactic". The density if the interstellar medium varies but is on the rough order of one molecule per cm³. The chance that a molecule will hit something as it passes through any particular cubic centimeter is rather small, but there are so very many cubic centimeters to pass through that it eventually will hit something. (Also note that if the interstellar gas is ionized, the ions don't need to hit each other perfectly in order to exchange some momentum).
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Henning MakholmAug 19 '12 at 0:57

4 Answers
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It is a form of wave front arising from the scatter of the explosion parts. In the interaction with the medium, the average energy and momentum of the original particles/fragments is diminished transferring them to the wave front.

When there is no medium the particles/fragments/gas following momentum conservation disperse linearly until they meet an obstacle. If they are in a gravitational field they will follow the corresponding paths prescribed by the field instead of linearly, and one would have to solve for those the equations.

Supernovae, discussed in a comment by @JerrySchirmer create their own medium by the enormous amount of matter they have, thus shock waves are generated. That is a different story. If there is no medium there is no shock wave.

I am not a chemist but my understanding of a chemical explosion is that there is a very fast release of energy and PRODUCTS. So I would say that the products are accelerated and expand. Since there is no atmosphere to work against, the dynamics is quite simple. A couple of years ago NASA used an impactor to study a comet. Maybe there is a model of the plume in the published papers.

I was just reading a sample of some random book that I came across ('Great Formulas Explained') and stumbled upon this formula that deals with 'Explosions' : Taylor-Sedov Formula . I then got reminded of this post on 'Stackexchange - Physics' and thought of posting here.

Here's a snippet pulled from this sample:

When a strong explosion takes place, a shock wave forms that
propagates in a spherical manner away from the explosion. The shock
front separates the air mass that is heated and compressed due to the
explosion from the undisturbed air.

Using the concept of similarity solutions, the physicists Taylor and
Sedov derived a simple formula that describes how the radius (in m) of
such a shock sphere grows with time (in s). To apply it, we need to
know two additional quantities: the energy of the explosion E (in J)
and the density of the surrounding air D (in Kg/m3). Here's the
formula:

r = 0.93 * (E/D) power 0.2 * t power 0.4

This formula is also used for modelling Supernova explosions. A search for 'Taylor Sedov formula for explosions' on any good search engine will lead you a good number of sources to read. Here are a couple of 'em: